1. Discrete Random Variables: The Binomial Distribution

2. Bernoulli’s trials
J. Bernoulli ( ) analyzed the idea of repeated independent trials for discrete random variables that had two possible outcomes: success or failure
In his notation he wrote that the probability of success is denoted by p and the probability of failure is denoted by q or 1-p

3. Binomial distribution
The binomial distribution is just n independent individual (Bernoulli) trials added up.
It is the number of “successes” in n trials.
The sum of the probabilities of all the independent trials totals 1.
We can define a ‘success’ as a ‘1’, and a failure as a ‘0’.

4. Binomial distribution
“x” is a binomial distribution if its probability function is:
Examples (note – success/failure could be switched!):
probability of success
probability of failure
Situation
x=1
(success)
x=0
(failure)
Coin toss to get heads
Turns up heads
Turns up tails
Rolling dice to get 1
Lands on 1
Lands on anything but 1
While testing a product, how many are found defective
Product is defective
Product is not defective

5. Binomial distribution
The binomial distribution is just n independent ‘Bernoulli trials’ added up
It requires that the trials be done “with replacement” ex. testing bulbs for defects:
Let’s say you make many light bulbs
Pick one at random, test for defect, put it back
Repeat several times
If there are many light bulbs, you do not have to replace (it won’t make a significant difference)
The result will be the binomial probability of a defective bulb (#defective total sample).

6. Binomial distribution - formula
Let’s figure out a binomial random variable’s probability function or formula
Suppose we are looking at a binomial with n=3 (ex. 3 coin flips; ‘heads’ is a ‘success’)
We will start with ‘all tails’ P(x=0):
Can happen only one way: 000
Which is: (1-p)(1-p)(1-p)
Simplified: (1-p)3

7. Binomial distribution - formula
Let’s figure out a binomial probability function (for n = 3)
This time we want 1 success plus 2 failures (ex. 1 heads + 2 tails, or P(x=1)):
This can happen three ways: 100, 010, 001
Which is: p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p
Simplified: 3p(1-p)2

8. Binomial distribution - formula
Let’s figure out a binomial probability function (for n = 3)
We want 2 ‘successes’ P(x=2):
Can happen three ways: 110, 011, 101, or…
pp(1-p)+(1-p)pp+p(1-p)p, which simplifies to..
3p2(1-p)

9. Binomial distribution - formula
Let’s figure out a binomial probability function (n = 3)
We want all 3 ‘successes’ P(x=3):
This can happen only one way: 111
Which we represent as: ppp
Which simplifies to: p3

10. Binomial distribution - formula
Let’s figure out a binomial probability function – in summary, for n = 3, we have:
P(x)
(Where x is the number of successes; ex. # of heads)
The sum of these expressions is the binomial distribution for n=3. The resulting equation is an example of the Binomial Theorem.

11. Binomial distribution - formula
A quick review of the Binomial Theorem:
If we use q for (1 – p), then…
p3 + 3p2q +3pq2 + q3 = (p + q)3
which is an example of the formula:
(a + b)n = ____________________
(if you forget it, check it in your text)

12. Binomial distribution - formula
Let’s figure out a binomial r.v.’s probability function (the quick way to compute the sum of the terms on the previous slide) - now here’s the formula…
In general, for a binomial:
(the # of x successes with probability p in n trials)

13. Binomial distribution - formula
Which can be written as:
or P(x) = nCx px(1 – p)n-x
This formula is often called the general term of the binomial distribution.

14. Expected Value
The expected value of a binomial distribu-tion equals the probability of success (p) for n trials:
E(X) also equals the sum of the probabilities in the binomial distribution.

15. Binomial distribution - Graph
Typical shape of a binomial distribution:
Symmetric, with total P(x) = 1
Note: this is a theoretical graph – how would an experimental one be different? P x

16. Binomial distribution - example
A realtor claims that he ‘closes the deal’ on a house sale 40% of the time.
This month, he closed 1 out of 10 deals.
How likely is his claim of 40% if he only completed 1/10 of his deals this month?

17. Binomial distribution - example
By using the binomial distribution function, it’s possible to check if his assessment of his abilities (i.e. 40% ‘closes’) is likely:
P(0 deals)=

18. Binomial distribution - example
So it seems pretty unlikely that his assess-ment of his abilities is right:
The probability of closing 1 or fewer deals out of 10 if (as he claims) he closes deals 40% of the time is less than 5% or less than 1/20.
What % of ‘closes’ do you think would have the highest probability in this distribution, if his claim was right?

19. Binomial distribution - example
Now see if you can determine the expected number of closings if he had 12 deals this month, assuming 40% success.
We need the values of n (= ___) and of p (= ____).
E(X) = np = ______ - this means that we would expect him to close about _____ deals, if his claim is correct [End of first example.]

20. Binomial Distribution – ex. 2
Alex Rios has a batting average of for the season. In last night’s game, he had 4 at bats. What are the chances he had 2 hits?
You try this one! First ask 3 questions…

21. Binomial Distribution – ex. 2
Is ‘getting a hit’ a discrete random variable?
Is this a Bernoulli trial? How would you
define a “success” and a “failure”?
Is each time at bat an independent event?
If you can answer ‘yes’ to the three questions above, then you can use the binomial distribution formula to answer the problem.

22. Binomial Distribution – ex. 2
First determine the following values:
The number of trials (Alex is at bat __ times) – this is the value of n
The probability of success (Alex’s average is ___) – this is p
The probability of ‘failure’: 1 – p = ___
The # of successes asked for (his chances of getting ___ hits) – this is x
Now you can use the formula:

23. Binomial Distribution – ex. 2
Put in the values from the previous screen, and discuss your answers. [pause here]
Did you get P(2) = 0.275?
Is 2 the most likely number of hits for Alex last night? How about 1 or 3?
P(0 or 1 or 2 or 3 or 4 hits) = _____?

24. Hypergeometric distribution
What happens if you have a situation in which the trials are not independent (this most often happens due to not replacing a selected item).
Each trial must result in success or failure, but the probability of success changes with each trial.

25. Hypergeometric distribution
Consider taking a sample from a population, and testing each member of the sample for defects.
Do this sampling without replacement.
As long as the sample is small compared to the population, this is close to binomial.
But if the sample is large compared to the population, this is a hypergeometric dist.

26. Hypergeometric dist. - formula
A hypergeometric distribution differs from binomial ones since it has dependent trials.
Probability of x successes in r dependent trials, with number of successes a out of a total of n possible outcomes:

27. Hypergeometric dist. - formula
The full version of this formula is:
Expected Value – the average probability of a success is the ratio of success overall (a/n) times r trials: